mosfet2Rev - c Copyright 2010. W. Marshall Leach, Jr.,...

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Unformatted text preview: c Copyright 2010. W. Marshall Leach, Jr., Professor, Georgia Institute of Technology, School of Electrical and Computer Engineering. The MOSFET Device Symbols Whereas the JFET has a diode junction between the gate and the channel, the metal-oxide semiconductor FET or MOSFET differs primarily in that it has an oxide insulating layer separating the gate and the channel. The circuit symbols are shown in Fig. 1. Each device has gate (G), drain (D), and source (S) terminals. Four of the symbols show an additional terminal called the body (B) which is not normally used as an input or an output. It connects to the drain-source channel through a diode junction. In discrete MOSFETs, the body lead is connected internally to the source. When this is the case, it is omitted on the symbol as shown in four of the MOSFET symbols. In integrated-circuit MOSFETs, the body usually connects to a dc power supply rail which reverse biases the body-channel junction. In the latter case, the so-called “body effect” must be accounted for when analyzing the circuit. Figure 1: MOSFET symbols. Device Equations The discussion here applies to the n-channel MOSFET. The equations apply to the p-channel device if the subscripts for the voltage between any two of the device terminals are reversed, e.g. vGS becomes vSG . The n-channel MOSFET is biased in the active mode or saturation region for vDS ≥ vGS − vT H , where vT H is the threshold voltage. This voltage is negative for the depletion-mode device and positive for the enhancement-mode device. It is a function of the body-source voltage and is given by vT H = VT O + γ φ − vBS − φ (1) where VT O is the value of vT H with vBS = 0, γ is the body threshold parameter, φ is the surface potential, and vBS is the body-source voltage. The drain current is given by iD = kW (1 + λvDS ) (vGS − vT H )2 2L (2) where W is the channel width, L is the channel length, λ is the channel-length modulation parameter, and k is given by ox k = µ0 Cox = µ (3) tox 1 In this equation, µ0 is the average carrier mobility, Cox is the gate oxide capacitance per unity area, ox is the permittivity of the oxide layer, and tox is its thickness. It is convenient to define a transconductance coefficient K given by kW K= (1 + λvDS ) = K0 (1 + λvDS ) (4) 2L where K0 is given by kW (5) K0 = 2L With these definitions, the drain current can be written iD = K (vGS − vT H )2 (6) Note that K plays the same role in the MOSFET drain current equation as β plays in the JFET drain current equation. Some texts define K = k (W/L) (1 + λvDS ) so that iD is written iD = (K/2) (vGS − vT H )2 . In this case, the numerical value of K is twice the value used here. To modify the equations given here to conform to this usage, replace K in any equation given here with K/2. Transfer and Output Characteristics The transfer characteristics are a plot of the drain current iD as a function of the gate-to-source voltage vGS with the drain-to-source voltage vDS held constant. Fig. 2 shows the typical transfer characteristics for a zero body-to-source voltage. In this case, the threshold voltage is a constant, i.e. vT H = VT O . For vGS ≤ VT O , the drain current is zero. For vGS > VT O , Eq. (6) shows that the drain current increases as the square of the gate-to-source voltage. The slope of the curve represents the small-signal transconductance gm , which is defined in the following. Figure 2: Drain current iD versus gate-to-source voltage vGS for constant drain-to-source voltage vDS . The output characteristics are a plot the drain current iD as a function of the drain-to-source voltage vDS with the gate-to-source voltage vGS and the body-to-source voltage vBS held constant. Fig. 3 shows the typical output characteristics for several values of gate-to-source voltage vGS . The dashed line divides the triode region from the saturation or active region. In the saturation region, the slope of the curves represents the reciprocal of the small-signal drain-source resistance r0 , which is defined in the next section. Small-Signal Models There are two small-signal circuit models which are commonly used to analyze MOSFET circuits. These are the hybrid-π model and the T model. The two models are equivalent and give identical results. They are described below. In addition, a simplified small-model is derived which is called the source equivalent circuit. The models are first developed for the case of no body effect and then with the body effect. The former 2 Figure 3: Drain current iD versus drain-to-source voltage vDS for constant gate-to-source voltage vGS . case assumes that the body-source voltage is zero, i.e. vBS = 0. This is the case with discrete MOSFETs in which the source is connected physically to the body. It also applies to small-signal ac analyses for which the body and source leads are connected to the same or different dc voltages. In this case, the small-signal body-source voltage is zero, i.e. vbs = 0, and there is no body effect. No Body Effect The small-signal models in this section assume that the body lead is connected to the source lead. The models also apply when the body and source leads are connected to different dc voltages so that the ac or signal voltage from body to source is zero. Hybrid-π Model Consider the case where the body-source voltage is zero, i.e. vBS = 0. In this case, the threshold voltage in Eq. 1 is a constant and given by vT H = VT O . Let the drain current and each voltage be written as the sum of a dc component and a small-signal ac component as follows: iD = I D + id (7) vGS = VGS + vgs (8) vDS = VDS + vds (9) If the ac components are sufficiently small, we can write id = ∂ID ∂ID vgs + vds ∂VGS ∂VDS (10) where the derivatives are evaluated at the dc bias values. Let us define gm = r0 = ∂ ID ∂VDS ∂ID = K (VGS − VT H ) = 2 K ID ∂VGS −1 = kW λ (VGS − VT H )2 2L −1 = 1/λ + VDS ID (11) (12) It follows that the small-signal drain current can be written id = id + 3 vds r0 (13) where id = gm vgs (14) The small-signal circuit which models these equations is given in Fig. 4(a). This is called the hybrid-π model. Figure 4: (a) Hybrid-π model. (b) T model. T Model The T model of the MOSFET is shown in Fig. 4(b). The resistor r0 is given by Eq. (12). The resistor rs is given by 1 rs = (15) gm where gm is the transconductance defined in Eq. (11). The currents are given by id = is + is = vds r0 (16) vgs = gm vgs rs (17) The currents in the T model are the same as for the hybrid-π model. Therefore, the two models are equivalent. Note that the gate and body currents in Fig. 4(b) are zero because the controlled source supplies the current that flows through rs . The Drain Equivalent Circuit If the FET output is taken from the drain, the input can be either applied to the gate or to the source. If it is applied to the gate, the circuit is called a common-source amplifier. If it is applied to the source, the circuit is called a common-gate amplifier. In some cases, separate inputs can be applied to both the gate and the source. In any of these cases, the drain output can be solved for by first making a small-signal Thévenin or Norton equivalent circuit seen looking into the drain. We solve for the Norton equivalent circuit here. We assume that the circuits external to the gate and the source can be represented by Thévenin equivalents. Figure 5(a) shows the FET symbol with separate Thévenin sources connected to the gate and the source. The bias circuits are not shown, but we assume that the bias solutions are known. Figure 5(b) shows the circuit with the FET replaced with the hybrid-π model. The Norton equivalent circuit seen looking into the drain consists of a parallel current source id(sc) and resistor rid connecting between the drain and ground. This is shown in Figure 5(c). The value of id(sc) is the drain current with vd = 0, i.e. with the drain node grounded. From Figure 5(b), this current is given by id(sc) = id + i0 4 id (18) Figure 5: (a) FET with Thevenin sources connected to the gate and the source. (b) Circuit with the FET replaced with its hybrid-π model. (c) Drain Norton equivalent circuit. where the approximation assumes that the current i0 through r0 is small compared to id . This is usually a very good approximation because r0 is a large value resistor. We call it the “r0 approximation” when the current i0 is neglected. In many cases, r0 is taken to be an infinite resistor, in which case the approximation is exact. To solve for id , we can write the loop equation vtg − vts = vgs + is Rts = vgs + (is + i0 ) Rts id = + (id + i0 ) Rts gm 1 id + Rts gm (19) where the relations vgs = id /gm and is = id have been used. It follows that we can write id(sc) = id = Gm (vtg − vts ) (20) where Gm is an equivalent transconductance given by Gm = 1 1 or = 1 rs + Rts + Rts gm (21) where rs = 1/gm . We next solve for the resistance rid seen looking into the drain node. Consider the drain current id to be an independent current source and set vtg = vts = 0. We can write vd = i0 r0 + is Rts = (id − id ) r0 + is Rts = id (r0 + Rts ) − id r0 id = gm vgs = −gm vs = −gm is Rts = −gm id Rts (22) (23) where vgs = −vs and is = id have been used. Substitution of id from the second equation into the first equation yields vd = id (r0 + Rts ) + gm id Rts r0 = id [r0 (1 + gm Rts ) + Rts ] 5 (24) It follows that the drain resistance is given by rid = vd Rts or = r0 (1 + gm Rts ) + Rts = r0 1 + id rs + Rts (25) Note that no approximations have been made in solving for rid . In summary, the small-signal Norton equivalent circuit seen looking into the drain of a FET is a current source id(sc) in parallel with a resistor rid given by id(sc) = id = Gm (vtg − vts ) Gm = (26) 1 1 or = 1 rs + Rts + Rts gm or rid = r0 (1 + gm Rts ) + Rts = r0 1 + Rts rs (27) + Rts (28) where vtg and vts , respectively, are the Thévenin voltages seen looking out of the gate and source and Rts is the Thévenin resistance in series with vts . Note that Rtg does not appear in the equations because the current through it is zero. Example 1 Figure 6(a) shows the signal equivalent circuit of a common-source amplifier. It is given that Rtg = 1 k , Rts = 50 , RD = 10 k , ID = 1 mA, K = 1.5 mA/ V2 , and r0 = 50 k . Solve for the voltage gain and output resistance of the circuit. Figure 6: (a) Common-source amplifier. (b) Common-gate amplifier. (c) Common-drain amplifier. √ Solution: gm = 2 KID = 2.45 mS, rs = 1/gm = 408 . A flow graph for the voltage gain is shown in Figure 7(a). From the flow graph, we can write vo i vo = d× = Gm × − (rid RD ) vtg vtg id The numerical values are Gm = 1 1 = rs + Rts 458 6 (29) (30) rid Rts + Rts rs 50 = 50k 1 + + 50 = 56.2 k 408 = r0 1 + 1 vo 56.2k × 10k = Gm × − (rid RC ) = ×− = −18.5 vtg 458 56.2k + 10k 56.2k × 10k = 8.49 k 56.2k + 10k Because the gain is negative, the amplifier is said to be an inverting amplifier. rout = rid RD = (31) (32) (33) Figure 7: (a) Flow graph for the CS amplifier. (b) Flow graph for the CG amplifier. Example 2 Figure 6(b) shows the signal equivalent circuit of a common-gate amplifier. It is given that Rts = 50 , RD = 10 k , ID = 1 mA, K = 1.5 mA/ V2 , and r0 = 50 k . Solve for the voltage gain and output resistance of the circuit. √ Solution: gm = 2 KID = 2.45 mS, rs = 1/gm = 408 . A flow graph for the voltage gain is shown in Figure 7(b). From the flow graph, we can write vo i vo = d× = −Gm × − (rid RC ) vts vts id The numerical values are Gm = rid 1 1 1 = = rs + Rts 408 + 50 458 Rts + Rts rs 50 = 50k 1 + + 50 = 56.2 k 408 = r0 1 + vo 1 56.2k × 10k = −Gm × − (rid RC ) = × = 18.5 vtg 458 56.2k + 10k 56.2k × 10k = 8.49 k 56.2k + 10k Because the gain is positive, the amplifier is said to be a non-inverting amplifier. rout = rid RD = The Gate Equivalent Circuit Because the gate current ig = 0, the equivalent circuit seen looking into the gate is an open circuit. 7 Figure 8: (a) BJT symbol with a Thévenin source connected to the base. (b) Circuit with the BJT replaced with its hybrid-π model. (c) Thévenin emitter equivalent circuit. The Source Equivalent Circuit Figure 8(a) shows the FET symbol with a Thévenin source connected to the gate. The bias circuits are not shown, but we assume that the bias solutions are known. We wish to solve for the small-signal Thévenin equivalent circuit seen looking into the source. Figure 8(b) shows the circuit with the FET replaced with the hybrid-π model. From the circuit in 8(b), we can write vs = vtg − vgs i = vtg − d gm i = vtg − s gm is − i0 = vtg − gm is vtg − gm (34) where the approximation assumes i0 is small compared to is has been used. It follows that the Thévenin equivalent circuit seen looking into the source is the voltage source vtg in series with a resistance ris given by 1 ris = = rs (35) gm The equivalent circuit is shown in Figure 8(c). There is no Rtg in this solution because the current through it is zero. With the definition of ris , we can define another way of calculating id(sc) in the Norton drain circuit. The current is in Figure 5(a) is given by vtg − vts is = (36) ris + Rts Because id = is is and id(sc) = id = Gm (vtg − vts ), we have a third equation for Gm given by Gm = 1 ris + Rts (37) Example 3 Figure 6(c) shows the signal equivalent circuit of a common-drain amplifier. It is given that Rtg = 10 k , Rts = 1 k , ID = 1 mA, K = 1.5 mA/ V2 , and r0 = 50 k . Solve for the voltage gain and output resistance of the circuit. 8 √ Solution: gm = 2 KID = 2.45 mS, rs = 1/gm = 408 . ris = rs = Gm = 1 = 408 gm 1 1 1 = = ris + Rts 408 + 1k 1408 The output resistance is rout = ris Rts = 408 × 1k = 290 408 + 1k The input resistance is an open circuit. Two possible flow graphs for the solution are shown in Figure 9. Figure 9: Flow graphs for the common-collector amplifier. The first solution for the voltage gain is illustrated in Figure 9(a), where voltage division is used to solve for the gain to obtain vo Rts 1k = = = 0.710 vtg ris + Rts 408 + 1k The second solution is illustrated in Figure 9(b). The voltage gain is vo i is vo 1 = d× × = Gm × 1 × Rts = × 1 × 1k = 0.710 vtg vtg id is 1408 Example 4 Figure 10(a) shows a CS/CD amplifier. What are the expressions for the input resistance, the output resistance, and the voltage gain? Solution: Because ig1 = 0, the input resistance is an open circuit, i.e. rin = ∞. The output resistance is rout = ris2 RS 2 where ris2 = rs2 = 1/gm2 . A flow graph for the voltage gain is shown in Figure 11(a). The gain is given by vo i vtg2 vo RS 2 = d1 × × = Gm1 × − (rid1 RD1 ) × vtg vtg id1 vtg2 ris2 + RS 2 where Gm1 = 1/ (ris1 + RS 1 ) and rid1 = r01 (1 + gm1 RS 1 ) + RS 1 . Example 5 Figure 10(b) shows a combination CS amplifier and a CD/CG amplifier. What are the expressions for the input resistances, the output resistance, and the output voltage? Solution: Because ig1 == ig2 = 0, both input resistances are open circuits, i.e. rin1 = rin2 = ∞. The output resistance is rout = rid1 RD1 where ris1 = RS + ris2 = RS + 1/gm2 . A flow graph for the output voltage is shown in Figure 12. It is given by vo = id1 × rid1 RD1 = Gm1 × (vtg1 − vts1 ) × rid1 RD1 = Gm1 × (vtg1 − vtg2 ) × rid1 RD1 9 Figure 10: (a) CS/CD amplifier. (b) Combination CS amplifier and CD/CG amplifier. Figure 11: Flow graph for the CS/CD amplifier. 10 where Gm1 = 1 = ris1 + RS + ris2 1 1 gm1 rid1 = r01 1 + gm1 RS + 1 gm2 + RS + 1 gm2 + RS + 1 gm2 Figure 12: Flow graph for the combination CS amplifier and CD/CG amplifier. The Body Effect The small-signal models above assume that the body lead is connected to the source lead. In the following, we assume that the body lead is connected to ac signal ground. In integrated circuit design, this ac signal ground is typically a dc power supply rail. In this case, any ac signal voltage on the source lead causes an ac signal voltage between the body and source. The effect of this voltage is called the body effect. Hybrid-π Model Let the drain current and each voltage be written as the sum of a dc component and a small-signal ac component as follows: iD = I D + id (38) vGS = VGS + vgs (39) vBS = VBS + vbs (40) vDS = VDS + vds (41) If the ac components are sufficiently small, we can write id = ∂ID ∂ID ∂ID vgs + vbs + vds ∂VGS ∂VBS ∂VDS (42) where the derivatives are evaluated at the dc bias values. Let us define ∂ID = K (VGS − VT H ) = 2 K ID ∂VGS √ ∂ID γ KID gmb = =√ = χgm ∂VBS φ − VBS γ χ= √ 2 φ − VBS gm = r0 = ∂ ID ∂VDS −1 = kW λ (VGS − VT H )2 2L 11 −1 = VDS + 1/λ ID (43) (44) (45) (46) The small-signal drain current can thus be written id = id + vds r0 id = idg + idb (47) where idg = gm vgs (48) idb = gmb vbs (49) The small-signal circuit which models these equations is given in Fig. 13(a). This is called the hybrid-π model. If the body (B) lead is connected to the source, then vbs = 0 and the circuit becomes that given in Fig. 4(a). Figure 13: (a) Hybrid-π model with body effect. (b) T model with body effect. T Model The T model of the MOSFET is shown in Fig. 4(b). The resistor r0 is given by Eq. (12). The resistors rs and rsb are given by 1 rs = (50) gm rsb = 1 1 rs = = gmb χgm χ (51) where gm and gmb are the transconductances defined in Eqs. (43) and (44). The currents are given by id = isg + isb + vds r0 vgs = gm vgs rs vbs = = gmb vbs rsb (52) isg = (53) isb (54) The currents are the same as for the hybrid-π model. Therefore, the two models are equivalent. Note that the gate and body currents are zero because the two controlled sources supply the currents that flow through rs and rsb . 12 A Simplified T Model There is a simplification to the T model with body effect that simplifies many calculations. Figure 14(a) shows the T model of the MOSFET with a Thévenin source connected to the gate and the body connected to signal ground. We desire the Thévenin equivalent circuit seen looking up into the is branch. The open circuit or Thévenin voltage is given by voc 1 vtg vtg rsb gmb = vtg = vtg = gmb = 1 + χ 1 1 rsg + rsb 1+ + gm gm gmb (55) The Thévenin resistance is calculated with vtg = 0. Let this be denoted by rs . It is given by rs = rs rsb = rs rsb rs or 1 = = rs + rsb 1+χ (1 + χ) gm (56) Figure 14(b) shows the simplified T model. For the case where the body and the source connect to the same signal node, set χ = 0 in the equations. Figure 14: (a) T model with a Thévenin source connected to the gate and the body grounded. (b) Simplified T mode. The Drain Equivalent Circuit Figure 15(a) shows the MOSFET with Thévenin sources connected to the gate and source leads. We wish to solve for the Norton equivalent circuit seen looking into the drain. The circuit consists of a parallel current source id(sc) and resistor rid connecting between the drain and ground. The value of id(sc) is the drain current with vd = 0, i.e. with the drain node grounded. From the simplified T model circuit in Figure 15(b), this current is given by id(sc) = id + i0 id (57) where the approximation assumes that the current i0 through r0 is small compared to id . This is usually a very good approximation because r0 is a large value resistor. We call it the “r0 approximation” when the current i0 is neglected. In many cases, r0 is taken to be an infinite resistor, in which case the approximation is exact. To solve for id , we can write the loop equation vtg − vts = is rs + is Rts = is rs + (is + i0 ) Rts = id rs + (id + i0 ) Rts 1+χ 13 id (rs + Rts ) (58) Figure 15: (a) MOSFET with Thévenin sources connected to the gate and the source. (b) Simplified T model of the circuit. (c) Norton equivalent drain circuit. From this equation, it follows that we can write id(sc) = id = Gm (vtg − vts ) (59) where Gm is an equivalent transconductance given by Gm = 1 or = rs + Rts 1 1 or = rs 1 + Rts + Rts 1+χ (1 + χ) gm (60) We next solve for the resistance rid seen looking into the drain node. Consider the drain current id to be an independent current source and set vtg = vts = 0. We can write vd = i0 r0 + is Rts = (id − id ) r0 + is Rts = id (r0 + Rts ) − id r0 id = is = (61) vgs is Rts id Rts =− =− rs rs rs where vgs = −vs = −is Rts and is = id have been used. Substitution of id from the second equation into the first equation yields vd = id (r0 + Rts ) − id r0 = id (r0 + Rts ) + id Rts Rts r0 = id r0 1 + rs rs + Rts (62) It follows that the drain resistance is given by rid = Rts vd = r0 1 + id rs or + Rts = r0 1 + (1 + χ) Rts or + Rts = r0 [1 + (1 + χ) gm Rts ] + Rts rs (63) Note that no approximations have been made in solving for rid . In summary, the small-signal Norton equivalent circuit seen looking into the drain of a FET is a current source id(sc) in parallel with a resistor rid given by id(sc) = id = Gm 14 vtg − vts 1+χ (64) Gm = rid = r0 1 + Rts rs 1 or = rs + Rts 1 1 or = rs 1 + Rts + Rts 1+χ (1 + χ) gm or + Rts = r0 1 + (1 + χ) Rts or + Rts = r0 [1 + (1 + χ) gm Rts ] + Rts rs (65) (66) where vtg and vts , respectively, are the Thévenin voltages seen looking out of the gate and source and Rts is the Thévenin resistance in series with vts . Note that Rtg does not appear in the equations because the current through it is zero. The Source Equivalent Circuit The source equivalent circuit is derived above for the case where the body lead is connected to the MOSFET gate. The circuit derived here is for the case where the body is connected to signal ground. Figure 16(a) shows the MOSFET symbol with a Thévenin equivalent source connected to the gate. The equation for the source voltage vs follows from the simplified T model circuit in Figure 14(b). It is given by vs = vtg vtg − is rs = − (is − i0 ) rs 1+χ 1+χ vtg − is rs 1+χ (67) where the approximation assumes that i0 is small compared to is . It follows that the source equivalent circuit consists of a voltage source vs(oc) in series with a resistance ris given by vs(oc) = vtg 1+χ ris = rs = 1 rs = 1+χ (1 + χ) gm (68) The circuit is shown in Figure 16(b). Figure 16: (a) MOSFET with Thevenin source connected to the gate and the body connected to signal ground. (b) Source equivalent circuit. Summary of Models id(sc) = id = Gm or Gm = 1 or = rs + Rts rs = 1 gm vtg − vts 1+χ 1 1 1 or or = = rs 1 ris + Rts + Rts + Rts 1+χ gm (1 + χ) ris = rs = 15 rs or 1 = 1+χ gm (1 + χ) (69) (70) (71) Figure 17: Summary of the equivalent circuits. rid = r0 1 + Rts rs Rts or + Rts = r0 [1 + (1 + χ) gm Rts ] + Rts rs or + Rts = r0 1 + (1 + χ) rig = ∞ (72) (73) Set χ = 0 when the body is connected to the source or when the body and the source are connected to the same node. Small-Signal High-Frequency Models Figures 18 and 19 show the hybrid-π and T models for the MOSFET with the gate-source capacitance cgs , the source-body capacitance csb , the drain-body capacitance cdb , the drain-gate capacitance cdg , and the gate-body capacitance cgb added. These capacitors model charge storage in the device which affect its high-frequency performance. The first three capacitors are given by 2 cgs = W LCox 3 csb = cdb = csb0 (1 + VSB /ψ0 )1/2 cdb0 (1 + VDB /ψ0 )1/2 (74) (75) (76) where VSB and VDB are dc bias voltages; csb0 and cdb0 are zero-bias values; and ψ0 is the built-in potential. Capacitors cgd and cgb model parasitic capacitances. For IC devices, cgd is typically in the range of 1 to 10 fF for small devices and cgb is in the range of 0.04 to 0.15 fF per square micron of interconnect. 16 Figure 18: High-frequency hybrid-π model. Figure 19: High-frequency T model. 17 ...
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This note was uploaded on 05/26/2011 for the course ECE 3050 taught by Professor Hollis during the Summer '08 term at Georgia Institute of Technology.

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mosfet2Rev - c Copyright 2010. W. Marshall Leach, Jr.,...

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